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CBSE Class 12 Math 2011 Solved Paper

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Question : 23
Total: 29
Using matrix method, solve the following system of equations:
2
x
+
3
y
+
10
z
 = 4 ,
4
x
−
6
y
+
5
z
 = 1 ,
6
x
+
9
y
−
20
z
 = 2 ; x , y , z ≠ 0 OR
Using elementary transformations, find the inverse of the matrix (
13−2
−30−1
210
)
Solution:  
The given system of equation is
2
x
+
3
y
+
10
z
 = 4 ,
4
x
−
6
y
+
5
z
 = 1 ,
6
x
+
9
y
−
20
z
 = 2
The given system of equation can be written as
[
2310
4−65
69−20
][
1
x
1
y
1
z
] = [
4
1
2
]
or AX B,Where A = [
2310
4−65
69−20
] , X = [
1
x
1
y
1
z
] and B = [
4
1
2
]
Now, |A| = [
2310
4−65
69−20
]
= 2 (120 - 45) - 3 (- 80 - 30) + 10 (36 + 36)
= 1200 ≠ 0
Hence, the unique solution of the system of equation is given by X = A−1B
Now, the cofactors of A are computed as :
C11 = (−1)2 (120 - 45) = 75, C12 = (−1)3 (- 80 - 30) = 110, C13 = (−1)4 (36 + 36) = 72
C21 = (−1)3 (- 60 - 90) = 150, C22 = (−1)4 (- 40 - 60) = - 100, C23 = (−1)5 (18 - 18) = 0
C31 = (−1)4 (15 + 60) = 75, C12 = (−1)5 (10 - 40) = 30, C33 = (−1)6 (- 12 - 12) = - 24
∴ Adj A = [
7511072
150−1000
7530−24
]T = [
7515075
110−10030
720−24
]
⇒ S−1 =
AdjA
|A|
 =
1
1200
[
7515075
110−10030
720−24
]
X = A−1B
=
1
1200
[
7515075
110−10030
720−24
][
4
1
2
]

=
1
1200
[
400+150+150
440−100+60
288+0−48
] =
1
1200
[
600
400
240
]
X = [
600
1200
400
1200
240
1200
] = [
1
2
1
3
1
5
] ⇒ [
1
x
1
y
1
z
] = [
1
2
1
3
1
5
]
⇒
1
x
 =
1
2
,
1
y
 =
1
3
 and
1
z
 =
1
5
⇒ x = 2 , y = 3 and z = 5
Thus, solution of given system of equation is given by x = 2, y = 3 and z = 5.
OR
The given matrix is A = [
13−2
−30−1
210
]
We have AA−1 = I
Thus, A = IA
Or, [
13−2
−30−1
210
] = [
100
010
001
] A
Applying R2 → R2+3R1 and R3 → R3−2R1
[
13−2
09−7
0−54
] = [
100
310
−201
] A
Now,applying R2 →
1
9
R2
[
13−2
01−
7
9
0−54
] = [
100
1
3
1
9
0
−201
] A
Applying R1 → R1−3R2 and R3 → R3+5R2
[
10
1
3
01−
7
9
00
1
9
] = [
0−
1
3
0
1
3
1
9
0
−
1
3
5
9
1
] A
Applying R3 → 9R3
[
10
1
3
01−
7
9
001
] = [
0−
1
3
0
1
3
1
9
0
−359
] A
Applying R1 → R1−
1
3
R3 and R2 → R2+
7
9
R3
[
100
010
001
] = [
1−2−3
−247
−359
] A ⇒ I = [
1−2−3
−247
−359
] A
∴ A−1 = [
1−2−3
−247
−359
]
Hence, inverse of the matrix A is [
1−2−3
−247
−359
]
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